- #1
Dave27
First of all, I will like to say that I'm teaching myself physics because I've never learned it at school. So, please excuse my ignorance and bear with me.
This is not really a homework problem but rather a problem I thought on my own given a situation in real life.
1. Homework Statement
I'm a recreational lifter, so I train and lift weights from time to time. For this problem I would just consider measurements in 2 dimensions or even just one if I consider the other is irrelevant. I have 2 squat stands about 1.5m tall, 6.4cm width, and a barbell about 1.8m long which weights about 6kg. I'll put this bar over the squat stands and put a 20kg plate on one end of the bar as shown in the picture:
Picture 1
As you can see I took some measurements. IRL the base of the stands are 2 buckets filled up with concrete (much wider) each weight about 30kg (I don't think this matters but I'm throwing it out just in case). I've also put point A and B as pivot points. The measurement that says 56.8cm starts exactly at the center of the bar.
I know for a fact that this system will collapse with the weight plate and bar going off to the right side. The question is how much distance I need to move the squat stand at point A for the system to be at equilibrium.
It should look something like this:
Picture 2
For this matter, wouldn't the force of the bar at the left side of point A increase? or is it rather just the torque decreasing because the distance between point A and the Forces on the right side is decreasing. Also, in picture 1, linearly the weight of the bar just cancels out with the stands but what happens with the weight plate? Obviously it doesn't make the bar go through the stands, so, should I view the bar and the plate as whole and just say both objects just exert a normal force on the stands and is really just the torque that would make it fall. I'm pretty confused.
T = F * d
d being the distance from the pivot point to the force being applied and F of course being the force due to gravity and T for torque.
Linearly also F = m * a
For the system to be at equilibrium the sum of all of the forces must equal 0 same for the torques.
Here's the thing, I don't know if I should take both points or just one (any) to solve the problem. I guess I just took point A and see where it leads me.
Let's say m1 = mass of the bar and m2 = mass of the plate
Considering pivot point A on the second picture.
On the right side the forces are applying a negative torque (clockwise)[/B]
T1 = m1g * D1
T2 = m2g * D2
On the left side only the weight of the bar is applying a positve torque but it would be
T3 = m1g * D1
T3 - ( T1 + T2 ) = 0
but T3 = T1 so
-T2 = 0 (?)
That can't be.
Obviously I'm missing an important concept here. Please help.
where D is the distance between each
PD: If you want to know my background on physics (which is non existant) I made a thread here: https://www.physicsforums.com/threads/hello-from-peru.927603/#post-5855350
This is not really a homework problem but rather a problem I thought on my own given a situation in real life.
1. Homework Statement
I'm a recreational lifter, so I train and lift weights from time to time. For this problem I would just consider measurements in 2 dimensions or even just one if I consider the other is irrelevant. I have 2 squat stands about 1.5m tall, 6.4cm width, and a barbell about 1.8m long which weights about 6kg. I'll put this bar over the squat stands and put a 20kg plate on one end of the bar as shown in the picture:
Picture 1
As you can see I took some measurements. IRL the base of the stands are 2 buckets filled up with concrete (much wider) each weight about 30kg (I don't think this matters but I'm throwing it out just in case). I've also put point A and B as pivot points. The measurement that says 56.8cm starts exactly at the center of the bar.
I know for a fact that this system will collapse with the weight plate and bar going off to the right side. The question is how much distance I need to move the squat stand at point A for the system to be at equilibrium.
It should look something like this:
Picture 2
For this matter, wouldn't the force of the bar at the left side of point A increase? or is it rather just the torque decreasing because the distance between point A and the Forces on the right side is decreasing. Also, in picture 1, linearly the weight of the bar just cancels out with the stands but what happens with the weight plate? Obviously it doesn't make the bar go through the stands, so, should I view the bar and the plate as whole and just say both objects just exert a normal force on the stands and is really just the torque that would make it fall. I'm pretty confused.
Homework Equations
T = F * d
d being the distance from the pivot point to the force being applied and F of course being the force due to gravity and T for torque.
Linearly also F = m * a
For the system to be at equilibrium the sum of all of the forces must equal 0 same for the torques.
The Attempt at a Solution
Here's the thing, I don't know if I should take both points or just one (any) to solve the problem. I guess I just took point A and see where it leads me.
Let's say m1 = mass of the bar and m2 = mass of the plate
Considering pivot point A on the second picture.
On the right side the forces are applying a negative torque (clockwise)[/B]
T1 = m1g * D1
T2 = m2g * D2
On the left side only the weight of the bar is applying a positve torque but it would be
T3 = m1g * D1
T3 - ( T1 + T2 ) = 0
but T3 = T1 so
-T2 = 0 (?)
That can't be.
Obviously I'm missing an important concept here. Please help.
where D is the distance between each
PD: If you want to know my background on physics (which is non existant) I made a thread here: https://www.physicsforums.com/threads/hello-from-peru.927603/#post-5855350